Is it possible to show that
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$
without using the Beta function
$$\text{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and the generalized Euler sum
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k),\quad q\ge 2\ ?$$
By integration by parts I found
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=\color{blue}{\int_0^1\frac{\ln^2(1-x)\ln x}{x}dx}$$
Setting $1-x\to x$ gives the blue integral again.
This integral seems tough under such restrictions. All approaches are welcome.
thanks.
Best Answer
Feynman trick works fine here:
Let
$$I=\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$$
$$I(a)=\int_0^1\frac{\ln^2x\ln(1-ax)}{1-x}dx,\quad I(1)=I,\quad I(0)=0$$
$$\Longrightarrow I'(a)=-\int_0^1\frac{x\ln^2x}{(1-x)(1-ax)}dx=2\frac{\text{Li}_3(a)}{a}+2\frac{\text{Li}_3(a)-\zeta(3)}{1-a}$$
$$\therefore I= 2\int_0^1\frac{\text{Li}_3(a)}{a}da+2\underbrace{\int_0^1\frac{\text{Li}_3(a)-\zeta(3)}{1-a}da}_{IBP}$$
$$=2\zeta(4)+2\int_0^1\frac{\ln(1-a)\text{Li}_2(a)}{a}da$$
$$=2\zeta(4)-\text{Li}_2^2(1)$$
$$=2\zeta(4)-\frac{5}{2}\zeta(4)=-\frac12\zeta(4)$$
Bonus:
I noticed that this trick works for only even powers of $\ln x$ and by following the same technique we find the generalization:
Some cases:
$$\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}dx=12\zeta^2(3)-18\zeta(6)$$
$$\int_0^1\frac{\ln^6x\ln(1-x)}{1-x}dx=720\zeta(3)\zeta(5)-900\zeta(8)$$
For the case of odd powers of $\ln x$, we will need to use Euler sum or Beta function.